To simplify the expression:

\[
\frac{1}{3^{5n}} \times 9^{n-1} \times 27^{n+1}
\]

### Step-by-step solution:

1. **Rewrite the terms using powers of 3:**

   - \( 9 = 3^2 \), so \( 9^{n-1} = (3^2)^{n-1} = 3^{2(n-1)} = 3^{2n-2} \).
   - \( 27 = 3^3 \), so \( 27^{n+1} = (3^3)^{n+1} = 3^{3(n+1)} = 3^{3n+3} \).

   Now the expression becomes:

   \[
   \frac{1}{3^{5n}} \times 3^{2n-2} \times 3^{3n+3}
   \]

2. **Combine the powers of 3:**

   Using the rule \( a^m \times a^n = a^{m+n} \), we can combine the exponents of 3.

   \[
   3^{-(5n)} \times 3^{2n-2} \times 3^{3n+3} = 3^{-(5n) + (2n-2) + (3n+3)}
   \]

3. **Simplify the exponent:**

   Combine the terms in the exponent:

   \[
   -(5n) + (2n-2) + (3n+3) = -5n + 2n + 3n - 2 + 3 = 0 + 1 = 1
   \]

   So, the expression simplifies to:

   \[
   3^1 = 3
   \]

### Final Answer:
The simplified expression is:

\[
3
\]

Would you like more details, or do you have any questions?

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Here are 5 related questions for further practice:

1. Simplify \( \frac{1}{2^{3n}} \times 4^{n-1} \times 16^{n+2} \).
2. Simplify \( \frac{1}{5^{4n}} \times 25^{n} \times 125^{n-1} \).
3. Express \( 8^{2n} \times \frac{1}{2^{6n}} \) in terms of powers of 2.
4. Simplify \( 2^{3x-2} \times 4^{x+1} \times \frac{1}{8^x} \).
5. If \( x^a \times x^b = x^c \), what is \( c \) in terms of \( a \) and \( b \)?

**Tip:** Remember that when multiplying terms with the same base, you add their exponents.

Learn how to simplify the expression 1/3^5n × 9^(n-1) × 27^(n+1) by applying exponentiation rules. This step-by-step solution explores key algebraic concepts, including powers of numbers and combining exponents. Ideal for students in Grades 9-11.

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